QUESTION IMAGE
Question
- for the following integrals, choose whether cartesian coordinates, cylindrical coordinates, or spherical coordinates would be the best way to find the integral, and then express the integral as an iterated integral where you have determined all of the bounds of integration and expressed the integrand in the new coordinates. you do not need to compute the integral! (in terms of the definition of \best,\ you can always avoid chopping the domains that i have given you as long as you are in the right coordinates and so you should.)
(a) \\( \omega:=\\{(x, y, z) \text { in the first octant }: x^{2}+y^{2} \leq 25, \text { and } z \leq 10\\} \\),
\\( \iiint_{\omega} z^{2} d v \\).
(b) \\( \omega:=\\{(x, y, z) \text { in the first octant }: x^{2}+y^{2}+z^{2} \leq 25\\} \\),
\\( \iiint_{\omega} z^{2} d v \\).
(c) \\( \omega:=\\{(x, y, z) \text { in the first octant }: x+y+z \leq 7\\} \\),
\\( \iiint_{\omega}\left(x^{2}+y^{2}\
ight) d v \\).
(d) \\( \omega:=\\{(x, y, z): 0 \leq z \leq 4 \text { and } x^{2}+y^{2}+z^{2} \leq 25\\} \\),
\\( \iiint_{\omega} z d v \\).
(e) \\( \omega:=\\{(x, y, z): \sqrt{x^{2}+y^{2}} \leq 4 \text { and } x^{2}+y^{2}+z^{2} \leq 25\\} \\),
\\( \iiint_{\omega} z^{2} d v \\).
(f) \\( \omega:=\\{(x, y, z):-\sqrt{x^{2}+y^{2}} \leq z, x \geq 0, \text { and } x^{2}+y^{2}+z^{2} \leq 144\\} \\),
\\( \iiint_{\omega} x d v \\).
(a)
Step1: Choose coordinate system
The region has \(x^{2}+y^{2}\leq25\) (a circular - like projection in the \(xy\) - plane) and \(z\) is simple. Cylindrical coordinates \((r,\theta,z)\) where \(x = r\cos\theta\), \(y = r\sin\theta\), \(z=z\) and \(dV=r\ dz\ dr\ d\theta\) are suitable. The first - octant implies \(0\leq\theta\leq\frac{\pi}{2}\), \(0\leq r\leq5\) (since \(x^{2}+y^{2}=r^{2}\leq25\)), and \(0\leq z\leq10\).
Step2: Express the integrand and integral
The integrand \(z^{2}\) remains \(z^{2}\) in cylindrical coordinates. The triple - integral \(\iiint_{\Omega}z^{2}dV=\int_{\theta = 0}^{\frac{\pi}{2}}\int_{r = 0}^{5}\int_{z = 0}^{10}z^{2}r\ dz\ dr\ d\theta\)
(b)
Step1: Choose coordinate system
The region \(x^{2}+y^{2}+z^{2}\leq25\) (a sphere) in the first octant. Spherical coordinates \((
ho,\varphi,\theta)\) where \(x=
ho\sin\varphi\cos\theta\), \(y =
ho\sin\varphi\sin\theta\), \(z=
ho\cos\varphi\) and \(dV=
ho^{2}\sin\varphi\ d
ho\ d\varphi\ d\theta\) are appropriate. For the first octant, \(0\leq\theta\leq\frac{\pi}{2}\), \(0\leq\varphi\leq\frac{\pi}{2}\), and \(0\leq
ho\leq5\) (since \(x^{2}+y^{2}+z^{2}=
ho^{2}\leq25\)).
Step2: Express the integrand and integral
The integrand \(z^{2}=
ho^{2}\cos^{2}\varphi\). The triple - integral \(\iiint_{\Omega}z^{2}dV=\int_{\theta=0}^{\frac{\pi}{2}}\int_{\varphi = 0}^{\frac{\pi}{2}}\int_{
ho = 0}^{5}(
ho^{2}\cos^{2}\varphi)
ho^{2}\sin\varphi\ d
ho\ d\varphi\ d\theta=\int_{\theta=0}^{\frac{\pi}{2}}\int_{\varphi = 0}^{\frac{\pi}{2}}\int_{
ho = 0}^{5}
ho^{4}\cos^{2}\varphi\sin\varphi\ d
ho\ d\varphi\ d\theta\)
(c)
Step1: Choose coordinate system
The plane \(x + y+z=7\) (Cartesian - like structure). Cartesian coordinates \((x,y,z)\) are suitable. First, find the limits. From \(x + y+z=7\), we get \(z=7-(x + y)\). In the first octant, \(z\geq0\), so \(y\leq7 - x\) and \(x\leq7\). The triple - integral \(\iiint_{\Omega}(x^{2}+y^{2})dV=\int_{x = 0}^{7}\int_{y = 0}^{7 - x}\int_{z = 0}^{7-(x + y)}(x^{2}+y^{2})dz\ dy\ dx\)
(d)
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(a) Cylindrical coordinates: \(\int_{\theta = 0}^{\frac{\pi}{2}}\int_{r = 0}^{5}\int_{z = 0}^{10}z^{2}r\ dz\ dr\ d\theta\)
(b) Spherical coordinates: \(\int_{\theta=0}^{\frac{\pi}{2}}\int_{\varphi = 0}^{\frac{\pi}{2}}\int_{
ho = 0}^{5}
ho^{4}\cos^{2}\varphi\sin\varphi\ d
ho\ d\varphi\ d\theta\)
(c) Cartesian coordinates: \(\int_{x = 0}^{7}\int_{y = 0}^{7 - x}\int_{z = 0}^{7-(x + y)}(x^{2}+y^{2})dz\ dy\ dx\)
(d) Cylindrical coordinates: \(\int_{\theta = 0}^{2\pi}\int_{z = 0}^{4}\int_{r = 0}^{\sqrt{25 - z^{2}}}zr\ dr\ dz\ d\theta\)
(e) Cylindrical coordinates: \(\int_{\theta = 0}^{2\pi}\int_{r = 0}^{4}\int_{z=-\sqrt{25 - r^{2}}}^{\sqrt{25 - r^{2}}}z^{2}r\ dz\ dr\ d\theta\)
(f) Cylindrical coordinates: \(\int_{\theta=0}^{\frac{\pi}{2}}\int_{r = 0}^{12}\int_{z=-r}^{\sqrt{144 - r^{2}}}r^{2}\cos\theta\ dz\ dr\ d\theta\)